Math Is Fun Forum

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

my proof valid??

Last edited by Dragonshade (2008-05-08 08:12:59)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: my proof valid??

ln 2 and ln 5 having a common factor doesn't make sense.  Common factors only apply to integers.  Furthermore, ln 5 = km is stating that ln 5 is an integer, which would be a contradiction.  But it is not valid to state that ln 5 = km.

With a bit of teasing, you can get it down to:

And conclude that n = 1 (why?).  Then 5^m = 2 is of course ridiculous.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: my proof valid??

You may have a bit of trouble concluding that n=1, if you are not allowed to assume some things about rationals and irrationals.  So instead, look at it like:

Conclude that this number is irrational if n is not 1. (Remember m and n are relatively prime)

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: my proof valid??

Your next step should be

The last equation can only be true (given that m and n are integers) if m = n = 0, which is a contradiction.

Last edited by JaneFairfax (2008-05-08 09:51:47)

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: my proof valid??

Oh yea, 5 and 2 are prime , so 5^n and 2^m would be relatively prime , Thx, Ricky and Jane